A father is three times as old as his son. In 12 years time, he will be twice as old as his son. Find the present ages of the father and the son.

Let the present age of father be x years and the present age of his son be y years. The present age of father is three times the age of the son. Thus, we have x = 3 y x - 3 y = 0 After 12 years, father’s age will be ( x + 12 ) years and son’s age will be ( y + 12 ) years. Thus using the given information, we have x + 12 = 2 ( y + 12 ) x + 12 = 2 y + 24 x - 2 y - 12 = 0 So, we have two equations x - 3 y = 0 x - 2 y - 12 = 0 By using cross-multiplication, we have x - 3 × - 12 - ( - 2 ) × 0 = - y 1 × - 12 - 1 × 0 = 1 1 × - 2 - 1 × ( - 3 ) x 36 - 0 = - y - 12 - 0 = 1 - 2 + 3 x 36 = y 12 = 1 1 x 36 = y 12 = 1 x = 36 , y = 12 Hence, the present age of father is 36 years and the present age of son is 12 years.

E M B I B E

MATHEMATICS CLASS X>Chapter 3 - Pair of Linear Equations in Two Variables>EXERCISE 3.9>Q 4

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10th CBSE

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Six years hence a man's age will be three times the age of his son and three years ago he was nine times as old as his son. Find their present ages.

Important Questions on Pair of Linear Equations in Two Variables

MEDIUM

10th CBSE

IMPORTANT

MATHEMATICS CLASS X>Chapter 3 - Pair of Linear Equations in Two Variables>EXERCISE 3.9>Q 5

Ten years ago, a father was twelve times as old as his son and ten years hence, he will be twice as old as his son will be then. Find their present ages.